Question

The current in an ideal, long solenoid is varied at a uniform rate of 0.01 As⁻¹. The solenoid has 2000 turns/m and its radius is 6.0 cm. (a) Consider a circle of radius 1.0 cm inside the solenoid with its axis coinciding with the axis of the solenoid. Write the change in the magnetic flux through this circle in 2.0 seconds. (b) Find the electric field induced at a point on the circumference of the circle. (c) Find the electric field induced at a point outside the solenoid at a distance 8.0 cm from its axis.

Answer 1

Given:
Rate of change of current in the solenoid,$\frac{di}{dt}$ = 0.01 A/s for 2s

$\therefore \frac{di}{dt}$ = 0.02 A/s
n = 2000 turns/m
R = 6.0 cm = 0.06 m
r = 1 cm = 0.01 m

(a) ϕ = BA

Area of the circle, A = π × 1 × 10⁻⁴
$∆i=\frac{\mathrm{d}i}{\mathrm{d}t}\times ∆t$ = (0.01 A/s) × 2 = 0.02 A/s
$\Rightarrow ∆\varphi ={\mathrm{\mu }}_{0}nA∆i$
Now,
Δϕ = (4$\mathrm{\pi }$ × 10⁻⁷) × (2 × 10³) × ($\mathrm{\pi }$ × 10⁻⁴) × (2 × 10⁻²)
= 16π² × 10⁻¹⁰ Wb
= 157.91 × 10⁻¹⁰ Wb
= 1.6 × 10⁻⁵ Wb
∴$\frac{d\mathrm{\varphi }}{dt}$ for 1s = 0.785 Wb

(b) The emf induced due to the change in the magnetic flux is given by

$$\begin{gathered} e=\frac{\mathrm{d\varphi }}{\mathrm{d}t} \\ \int Edl=\frac{\mathrm{d\varphi }}{\mathrm{d}t} \\ \Rightarrow E\times 2\mathrm{\pi }r\mathit{=}\frac{\mathrm{d}\varphi }{\mathrm{d}t} \end{gathered}$$
The electric field induced at the point on the circumference of the circle is given by

$$\begin{gathered} E=\frac{0.785\times {10}^{-8}}{2\mathrm{\pi }\times {10}^{-2}} \\ =1.2\times {10}^{-7}\mathrm{V}/\mathrm{m} \end{gathered}$$

(c) For the point located outside the solenoid,

$$\begin{gathered} \frac{\mathrm{d}\varphi }{\mathrm{d}t}={\mathrm{\mu }}_{0}n\frac{\mathrm{d}i}{\mathrm{d}t}\mathrm{A} \\ =(4\mathrm{\pi }\times {10}^{-7})\times \left(2000\right)\times (0.01)\times (\mathrm{\pi }\times (0.06{)}^2) \\ E.dl=\frac{\mathrm{d}\varphi }{\mathrm{d}t} \\ \Rightarrow E=\frac{\mathrm{d}\varphi /\mathrm{d}t}{2\pi r} \end{gathered}$$

The electric field induced at a point outside the solenoid at a distance of 8.0 cm from the axis is given by
$$\begin{gathered} E=\frac{(4\mathrm{\pi }\times {10}^{-7})\times \left(2000\right)\times (0.01\times \mathrm{\pi }\times (0.06{)}^2)}{(\mathrm{\pi }\times 8\times {10}^{-2})}\times \frac{d\mathrm{B}}{dt} \\ =5.64\times {10}^{-7}\mathrm{V}/\mathrm{m} \end{gathered}$$